1. **State the problem:** Solve the differential equation $$c^2 (y')^2 = y^2 (y^{2n} - c^2)$$ with initial conditions $$y(a) = A$$ and $$y'(0) = 0$$. 2. **Rewrite the equation:** We have $$c^2 (y')^2 = y^2 (y^{2n} - c^2)$$ which implies $$y' = \pm \frac{y}{c} \sqrt{y^{2n} - c^2}$$. 3. **Separate variables:** Assuming $y > 0$ for simplicity, $$\frac{dy}{dx} = \pm \frac{y}{c} \sqrt{y^{2n} - c^2}$$ so $$\frac{dy}{y \sqrt{y^{2n} - c^2}} = \pm \frac{dx}{c}$$. 4. **Integrate both sides:** $$\int \frac{dy}{y \sqrt{y^{2n} - c^2}} = \pm \frac{1}{c} \int dx = \pm \frac{x}{c} + K$$ where $K$ is the integration constant. 5. **Substitution for integral:** Let $z = y^n$, then $dz = n y^{n-1} dy$, so $$dy = \frac{dz}{n y^{n-1}} = \frac{dz}{n z^{(n-1)/n}}$$. Rewrite the integral: $$\int \frac{dy}{y \sqrt{y^{2n} - c^2}} = \int \frac{1}{y \sqrt{z^2 - c^2}} dy = \int \frac{1}{y \sqrt{z^2 - c^2}} \cdot \frac{dz}{n z^{(n-1)/n}}$$ Since $y = z^{1/n}$, then $y = z^{1/n}$ and $y = z^{1/n}$, so $$\frac{1}{y} = z^{-1/n}$$ Thus the integral becomes $$\int \frac{z^{-1/n}}{\sqrt{z^2 - c^2}} \cdot \frac{dz}{n z^{(n-1)/n}} = \int \frac{1}{n \sqrt{z^2 - c^2}} dz = \frac{1}{n} \int \frac{dz}{\sqrt{z^2 - c^2}}$$. 6. **Integral evaluation:** $$\int \frac{dz}{\sqrt{z^2 - c^2}} = \ln \left| z + \sqrt{z^2 - c^2} \right| + C$$ 7. **Putting it all together:** $$\frac{1}{n} \ln \left| y^n + \sqrt{y^{2n} - c^2} \right| = \pm \frac{x}{c} + K$$ 8. **Solve for $y$:** Exponentiate both sides: $$\left| y^n + \sqrt{y^{2n} - c^2} \right| = e^{n(\pm \frac{x}{c} + K)} = Ce^{\pm \frac{n}{c} x}$$ where $C = e^{nK} > 0$. 9. **Apply initial condition $y(a) = A$:** $$\left| A^n + \sqrt{A^{2n} - c^2} \right| = C e^{\pm \frac{n}{c} a}$$ so $$C = \left| A^n + \sqrt{A^{2n} - c^2} \right| e^{\mp \frac{n}{c} a}$$ 10. **Final implicit solution:** $$\left| y^n + \sqrt{y^{2n} - c^2} \right| = \left| A^n + \sqrt{A^{2n} - c^2} \right| e^{\pm \frac{n}{c} (x - a)}$$ 11. **Check $y'(0) = 0$ condition:** From the original equation, $$y'(0) = 0 \implies y^{2n}(0) = c^2$$ so $$y(0) = c^{1/n}$$ which may impose restrictions on $A$, $a$, and the sign choice. **Summary:** The implicit solution is $$\left| y^n + \sqrt{y^{2n} - c^2} \right| = \left| A^n + \sqrt{A^{2n} - c^2} \right| e^{\pm \frac{n}{c} (x - a)}$$ with the initial conditions used to determine constants and sign.